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Maximize the volume without using Lagrange multipliers

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s3a
#1
Feb28-12, 09:01 AM
P: 584
1. The problem statement, all variables and given/known data
When a rectangular box is sent through the mail, the post office demands that the length of the box plus twice the sum of its height and width be no more than 250 centimeters. Find the dimensions of the box satisfying this requirement that encloses the largest possible volume. (Solve this problem without using Lagrange multipliers.)


2. Relevant equations
Partial differentiation and equations of constraint for each variable.


3. The attempt at a solution
My attempt at a solution is attached however, given that I got length = l = 0 (even though I get a nonzero width and height) which gives a volume of 0, I'd say I did something wrong and I don't have the solutions or answer for this particular problem so I can't check what's wrong.
Attached Thumbnails
MyWork.jpg  
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lanedance
#2
Feb28-12, 09:51 AM
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P: 3,307
did you just equate the volume with zero? shouldn't you differentiate to maximise the volume and find where the derivative is zero?
SammyS
#3
Feb28-12, 09:52 AM
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Quote Quote by s3a View Post
1. The problem statement, all variables and given/known data
When a rectangular box is sent through the mail, the post office demands that the length of the box plus twice the sum of its height and width be no more than 250 centimeters. Find the dimensions of the box satisfying this requirement that encloses the largest possible volume. (Solve this problem without using Lagrange multipliers.)

2. Relevant equations
Partial differentiation and equations of constraint for each variable.

3. The attempt at a solution
My attempt at a solution is attached however, given that I got length = l = 0 (even though I get a nonzero width and height) which gives a volume of 0, I'd say I did something wrong and I don't have the solutions or answer for this particular problem so I can't check what's wrong.


(Your solution is correct for the minimum volume.)

Once you have w = h, put that back into the volume formula so V is only a function of w or h .

Maximize that.

s3a
#4
Mar1-12, 01:02 PM
P: 584
Maximize the volume without using Lagrange multipliers

lanedance, no that's not what I did. I did differentiate (partially) hence the subscripts. ;)

SammyS, it makes sense that I got the minimum :) (thanks for mentioning it though because I initially thought I was doing something redundant rather than getting a minimum). What you said sounds like what I did though. Could you please be a bit more descriptive algorithmically?
SammyS
#5
Mar1-12, 07:00 PM
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Quote Quote by s3a View Post
SammyS, it makes sense that I got the minimum :) (thanks for mentioning it though because I initially thought I was doing something redundant rather than getting a minimum). What you said sounds like what I did though. Could you please be a bit more descriptive algorithmically?
[itex]\displaystyle V=\ell hw[/itex]

If h = w , then = 250 - 4h ,

and [itex]\displaystyle V=(250 - 4h)h^2\,.[/itex]

Maximize that.
s3a
#6
Mar1-12, 09:01 PM
P: 584
In the attachment above, I already have

V = 250w^2 - 4w^3

which is basically what you said with w instead of h which is okay since w = h.

As for maximizing that, do you mean taking dV/dw = 0 and solving for w?
SammyS
#7
Mar1-12, 09:56 PM
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Quote Quote by s3a View Post
In the attachment above, I already have

V = 250w^2 - 4w^3

which is basically what you said with w instead of h which is okay since w = h.

As for maximizing that, do you mean taking dV/dw = 0 and solving for w?
That's the usual way.

Try it.
s3a
#8
Mar1-12, 11:51 PM
P: 584
Is this what you meant?

(By the way, I choose to reject w = 0 since it yields a minimum.)
Attached Thumbnails
MyWork.jpg  


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